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August 21, 2014

August 21, 2014

Posted by **Rae** on Monday, July 2, 2012 at 3:00pm.

- Calculus -
**Reiny**, Monday, July 2, 2012 at 3:32pmThe two hemispheres at the ends will make up one complete sphere

Let the radius of the sphere be r ft, and let the height of the cylindrical part be h

Volume = cylinder + sphere

πr^2 h + (4/3)πr^3 = 144π

3πr^2 h + 4πr^3 = 432π

3r^2 h + 4r^3 = 432

h = (432 - 4r^3)/(3r^2)

The cost of production is dependent on the surface area of material used

SA = 4π^2 + 2πrh

= 4πr^2 + 2π(432 - 4r^3)/(3r^2)

= 4πr^2 + 288π/r^2 - (8/3)πr

d(SA)/dr = 8πr - 576π/r^3 - 8π/3 = 0 for a min of SA

8r - 576/r^3 - 8/3 = 0

times 3r^3

24r^4 -1728 - 8r^3 = 0

3r^4 - r^3 - 216 = 0

At this point I "cheated" a bit and ran it through Wolfram

http://www.wolframalpha.com/input/?i=3r%5E4+-+r%5E3+-+216+%3D+0

for a real solution of r=3, r = appr-3 and 2 complex roots,

so the only usable answer is r = 3

the h = (432 - 108)/27 = 12

So the cylinder part should be 12 ft long and the radius of each of the semispheres should be 3 ft

- Calculus -
**Damon**, Monday, July 2, 2012 at 5:22pmremember the sphere surface is twice as expensive as the cylinder which has no compound curvature so is easy to roll.

c = 2(4 pi r^2) + 1(2 pi r h)

c= 8 pi r^2 + 2 pi r h

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